[mercury-users] Type declaration for a higher order function without arguments?

[Fergus Henderson]

On 06-Aug-2004, Peter Hawkins <peter at hawkins.emu.id.au> wrote:
> Hi...
> 
> How does one declare a function type that doesn't take arguments?
> 
> This is ok:
> :- type myfunc(T) = (impure func(int) = T).
> 
> This is not:
> :- type myfunc(T) = (impure func = T).
> 
> Nor is this:
> :- type myfunc(T) = (impure func() = T).
> 
> What's the secret?

'func' is a prefix operator, so if you want to use it without
an operand, you need to enclose it in parentheses:

	:- type myfunc(T) == (impure (func) = T).

-- 
Fergus Henderson                    |  "I have always known that the pursuit
                                    |  of excellence is a lethal habit"
WWW: <http://www.cs.mu.oz.au/~fjh>  |     -- the last words of T. S. Garp.
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